Answer
(a) There are 2 chances to jump on the bus; first at 3.0 seconds and then at 6 seconds.
(b) The person can wait a maximum time of 0.25 seconds.
Work Step by Step
(a) The distance covered when running must be 9 meters more than the bus moves;
$v~t=\frac{1}{2}at^2+9$
$(4.5~m/s)~t=\frac{1}{2}(1.0~m/s^2)~t^2+9.0~m$
$(1.0~m/s^2)~t^2-(9.0~m/s)~t+18.0~m=0$
$(t-3.0)(t-6.0)=0$
$t = 3.0~s, 6.0~s$
There are 2 chances to jump on the bus; first at 3.0 seconds and then at 6 seconds.
(b) If we wait until the last possible moment, the bus's speed when we reach the door will be 4.5 m/s. We can find the time $t_b$ that the bus is moving.
$(1.0~m/s^2)~t_b = 4.5~m/s$
$t_b = 4.5 ~s$
We can find the distance $x$ the bus moves in this time.
$x = \frac{1}{2}at_b^2$
$x = \frac{1}{2}(1.0~m/s^2)(4.5~s)^2$
$x = 10.125~m$
The person needs to run a total distance of 19.125 meters. We can find the time $t_p$ it takes the person to run this distance.
$t_p = \frac{19.125~m}{4.5~m/s}$
$t_p = 4.25~s$
The person can wait a maximum time of 4.5 s - 4.25 s which is 0.25 seconds.
